Question

A narrow monochromatic beam of light of intensity $I$ is incidents on a glass plate as shown, in the figure. Another identical glass plate is kept close to the first-one and parallel to it. Each glass plate reflects $25\%$ of the light incident on it and transmits the remaining. Then, the ratio of the maximum and minimum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate is

• A. $7:1$
• B. $49:1$
• C. $4:1$
• D. $16:9$

Answer 1

The correct option is B $49:1$

As given that at each reflection 2525% of incident energy is reflected, therefore at each refraction 7575% of incident energy will be refracted.

Intensity of incident light =I=I

Hence, Intensity of reflected ray ABAB

$I_{AB}=25I/100=I/4$....................eq1

Intensity of refracted ray ACAC

$I_{AC}=75I/100=3I/4$
IAC=75I100=3I4

Intensity of reflected ray CA′CA′

$I_{C{A}^{\prime }}=25I_{AC}/100=1/4\times 3I/4=3I/16$

Intensity of refracted ray A′B′A′B′

$I_{A^{\prime }{B}^{\prime }}=75I_{C{A}^{\prime }}/100=3/4\times 3I/16=9I/64$ .......................eq2

Now, rays AB and A'B' interfere, therefore

$\frac{I_{max}}{I_{min}}=\frac{I_{AB}+I_{A^{\prime }{B}^{\prime }}+2\sqrt{I_{AB}{I}_{A^{\prime }{B}^{\prime }}}}{I_{AB}+I_{A^{\prime }{B}^{\prime }}-2\sqrt{I_{AB}{I}_{A^{\prime }{B}^{\prime }}}}$

$\frac{I_{max}}{I_{min}}=\frac{I/4+9I/64+2\sqrt{I/4\times 9I/64}}{I/4+9I/64-2\sqrt{I/4\times 9I/64}}=\frac{49}{1}$